Question

How do you find the point of intersection(s) for x = 2y2 + 3y + 1 and 2x + 3y2 = 0

A) You cannot find points of intersections for non-functions.

B) Plug in 0 for x into both equations and solve for y. Then plug that answer back into the other equation to find the corresponding x-coordinate.

C) Solve both equations for x and set them equal to each other. This will give you the y-coordinates of the points of intersection. Then plug back into one of the equations to find the corresponding x-coordinates.

D) Solve both equations for y and set them equal to each other. This will giv

Answer 1

You use method C, or method D. They are both technically the same method. One you find x first, the other you find y first.

(to say to the power of, we write ^(number)...)

x = 2y^2 + 3y + 1
and
2x + 3y2 =0
x=-3y^2/2

2y^2+3y+1=-3y^2/2
4y^2+6y+2=-3y^2

7y^2+6y+2=0

You have a quadratic. Solve it. You'll get TWO y values (this means the curves intersect at two places). Find the two y values, then sub BOTH y values (at different times) into the original formula and you'll get the corresponding x values.